Question

I was just wondering how disastrous integer overflow really is. Take the following example program:

#include <iostream>

int main()
{
    int a = 46341;
    int b = a * a;
    std::cout << "hello world\n";
}

Since a * a overflows on 32 bit platforms, and integer overflow triggers undefined behavior, do I have any guarantees at all that hello world will actually appear on my screen?

---

I removed the "signed" part from my question based on the following standard quotes:

(§5/5 C++03, §5/4 C++11) If during the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined.

(§3.9.1/4) Unsigned integers, declared unsigned, shall obey the laws of arithmetic modulo 2^n where n is the number of bits in the value representation of that particular size of integer. This implies that unsigned arithmetic does not overflow because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting unsigned integer type.

Answer 1

As pointed out by @Xeo in the comments (I actually brought it up in the C++ chat first):

Undefined behavior really means it and it can hit you when you least expect it.

The best example of this is here: Why does integer overflow on x86 with GCC cause an infinite loop?

On x86, signed integer overflow is just a simple wrap-around. So normally, you’d expect the same thing to happen in C or C++. However, the compiler can intervene – and use undefined behavior as an opportunity to optimize.

In the example taken from that question:

#include <iostream>
using namespace std;

int main(){
    int i = 0x10000000;

    int c = 0;
    do{
        c++;
        i += i;
        cout << i << endl;
    }while (i > 0);

    cout << c << endl;
    return 0;
}

When compiled with GCC, GCC optimizes out the loop test and makes this an infinite loop.